← Back to DSA Animator
Reorder List LC #143 Medium Linked List · 3-Phase
Problem

You are given the head of a singly linked-list: L₀ → L₁ → … → Lₙ₋₁ → Lₙ. Reorder it to: L₀ → Lₙ → L₁ → Lₙ₋₁ → L₂ → Lₙ₋₂ → …. You may not modify the values in the list's nodes — only the nodes themselves may be changed.

Example 1
Input: [1,2,3,4]
Output: [1,4,2,3]
Example 2
Input: [1,2,3,4,5]
Output: [1,5,2,4,3]
Constraints: 1 ≤ n ≤ 5×10⁴  |  1 ≤ Node.val ≤ 1000
Try Examples
Custom:
List Visualization
 Phase 1 — Find Middle
State Variables
Phase
—
Head Ptr (l1)
—
Tail Ptr (l2)
—
Step Logic
Press ▶ Play or Next Step to begin the animation.
🎉
Ready
0 / 0
Select an example above and press Play.
Algorithm
1
Phase 1: Use slow/fast pointers to find the middle node
2
Phase 2: Reverse the second half (slow.next … end)
3
Phase 3: Interleave l1 and l2 by rewiring .next pointers
4
Done — list is reordered in-place, O(n) time O(1) space
Time
O(n)
Space
O(1)
Why It Works

Finding the middle splits the list into two halves of equal (or near-equal) length. Reversing the second half lets us pair elements from both ends simultaneously. The interleave step then weaves them together by advancing one pointer from the front and one from the back, alternating which chain we stitch in each iteration — producing exactly the required ordering without any extra memory.